Why does int_a^(ab)1/xdx=int_1^b1/t dt? I can't understand how the integral having limits from a to ab in Step 1 is equivalent to the integral having limits from 1 to b. I'm a beginner here. Please explain in detail. ln(ab)=int_(1)^(ab)1/x dx = int_(1)^(a)(1)/(x)dx+int_(a)^(ab) 1/x dx =int_(1)^(a) 1/x dx +int_(1)^(b)1/(at) d(at)= int_(1)^(a)1/x dx +int_(1)^(b)1/t dt=ln(a)+ln(b)

imchasou 2022-09-20 Answered
Why does a a b 1 x d x = 1 b 1 t d t?
I can't understand how the integral having limits from a to a b in Step 1 is equivalent to the integral having limits from 1 to b. I'm a beginner here. Please explain in detail.
ln ( a b ) = 1 a b 1 x d x = 1 a 1 x d x + a a b 1 x d x = 1 a 1 x d x + 1 b 1 a t d ( a t ) = 1 a 1 x d x + 1 b 1 t d t = ln ( a ) + ln ( b ) .
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Answers (2)

Claire Larson
Answered 2022-09-21 Author has 10 answers
The substitution method was already given so take this method:
Let
f ( x ) = 1 x d t t
and
g ( x ) = a a x d t t
then we have
f ( x ) = g ( x ) = 1 x
and
f ( 1 ) = g ( 1 ) = 0
hence we conclude that
f ( x ) = g ( x ) , x R
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kennadiceKesezt
Answered 2022-09-22 Author has 2 answers
Your are asking why
a a b 1 x d x = 1 b 1 t d t
Well, it follows by substituting x = a t d x = a d t. Now, the limits in the first integral are x = a to x = a b. Hence, if x = a, then t = a a = 1. and if x = a b , then t = a b a = b which are your new limits of integration.
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